Related Questions

The simply supported beam 'A' of length '$$l$$' carries a central point load 'W'. Another beam 'B' is loaded with a uniformly distributed load such that the total load on the beam is 'W'. The ratio of maximum deflections between beams 'A' and 'B' is
A $$\frac{5}{8}$$
B $$\frac{8}{5}$$
C $$\frac{5}{4}$$
D $$\frac{4}{5}$$
The natural frequency of free transverse vibrations due to uniformly distributed load acting over a simply supported shaft is (where $$\delta {\text{S}}$$ = Static deflection of simply supported shaft due to uniformly distributed load)
A $$\frac{{0.4985}}{{\sqrt {\delta {\text{S}}} }}$$
B $$\frac{{0.5615}}{{\sqrt {\delta {\text{S}}} }}$$
C $$\frac{{0.571}}{{\sqrt {\delta {\text{S}}} }}$$
D $$\frac{{0.6253}}{{\sqrt {\delta {\text{S}}} }}$$
A simply supported beam A carries a point load at its mid span. Another identical beam B carries the same load but uniformly distributed over the entire span. The ratio of the maximum deflections of the beams A and B, will be
A $$\frac{2}{3}$$
B $$\frac{3}{2}$$
C $$\frac{5}{8}$$
D $$\frac{8}{5}$$
A simply supported beam 'A' of length l, breadth b, and depth d carries a central point load W. Another beam 'B' has the same length and depth but its breadth is doubled. The deflection of beam 'B' will be __________ as compared to beam 'A'.
A One-fourth
B One-half
C Double
D Four times
When a uniformly distributed load, longer than the span of the girder, moves from left to right, then the maximum bending moment at mid section of span occurs when the uniformly distributed load occupies
A Less than the left half span
B Whole of left half span
C More than the left half span
D Whole span
The natural frequency of free transverse vibrations due to a point load acting over a simply supported shaft is equal to (where $$\delta $$ = Static deflection of a simply supported shaft due to the point load)
A $$\frac{{0.4985}}{{\sqrt \delta }}$$
B $$\frac{{0.5615}}{{\sqrt \delta }}$$
C $$\frac{{0.571}}{{\sqrt \delta }}$$
D $$\frac{{0.6253}}{{\sqrt \delta }}$$
The bending moment in the center of a simply supported beam carrying a uniformly distributed load of w per unit length is
A Zero
B $$\frac{{{\text{w}}{l^2}}}{2}$$
C $$\frac{{{\text{w}}{l^2}}}{4}$$
D $$\frac{{{\text{w}}{l^2}}}{8}$$
The shear force at the ends of a simply supported beam carrying a uniformly distributed load of w per unit length is
A Zero at its both ends
B $${\text{w}}l$$ at one end and $$ - {\text{w}}l$$  at the other end
C $$\frac{{{\text{w}}l}}{2}$$ at one end and $$ - \frac{{{\text{w}}l}}{2}$$  at the other end
D $$\frac{{{\text{w}}{l^2}}}{2}$$ at one end and $$ - \frac{{{\text{w}}{l^2}}}{2}$$  at the other end
The shear force diagram for a simply supported beam carrying a uniformly distributed load of w per unit length, consists of
A One right angled triangle
B Two right angled triangles
C One equilateral triangle
D Two equilateral triangles
In a simply supported beam carrying a uniformly distributed load w per unit length, the point of contraflexure
A Lies in the center of the beam
B Lies at the ends of the beam
C Depends upon the length of beam
D Does not exist

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